Question

\[\mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}{{\sqrt {\left( {x + 1} \right)} }}\] , then the limit is equal to :
A. $\dfrac{1}{{\sqrt \pi }}$
B. $\dfrac{1}{{\sqrt {2\pi } }}$
C. 1
D. 0

Answer 1

Hint: This problem deals with solving the limit with L’Hospital’s rule. The L’Hospital’s rule is applied to a limit when the limit is in indeterminate form. This is done by differentiating the numerator and the denominator and then limit is applied again, which is given by:
$ \Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{f'(a)}}{{g'(a)}}$
Also the value of cosine trigonometric at $\pi $ radians is equal to -1:
$ \Rightarrow \cos \pi = - 1$
The derivative of ${\cos ^{ - 1}}x$ is $\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}$.

Complete step-by-step answer:
Using the L’Hospital’s rule to the given limit.
Consider the given limit, as given below:
\[ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}{{\sqrt {\left( {x + 1} \right)} }}\]
We know that the of $\cos \pi $, when is equal to $ - 1$, which is given below:
$ \Rightarrow \cos \pi = - 1$
Now taking the inverse of cosine on both sides of the equation, as given below:
$ \Rightarrow \pi = {\cos ^{ - 1}}\left( { - 1} \right)$
$\therefore {\cos ^{ - 1}}\left( { - 1} \right) = \pi $
That is when $x$ tends to -1 , then the value of the cosine inverse of $x$ is equal to $\pi $, as shown:
$ \Rightarrow \mathop {\lim }\limits_{x \to - 1} {\cos ^{ - 1}}x = \pi $
Now consider the limit of the numerator when $x$ tends to -1 of the given limit \[\mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}{{\sqrt {\left( {x + 1} \right)} }}\]as shown:
\[ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} \]
\[ \Rightarrow \sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}\left( { - 1} \right)} \right)} \]
As we know that $\because {\cos ^{ - 1}}\left( { - 1} \right) = \pi $, hence substituting this value as shown:
\[ \Rightarrow \sqrt \pi - \sqrt \pi = 0\]
So the value of the numerator is zero, when put the value of $x \to - 1$.
Now consider the limit of the denominator of the given limit \[\mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}{{\sqrt {\left( {x + 1} \right)} }}\] as shown:
$ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \sqrt {\left( {x + 1} \right)} $
$ \Rightarrow \sqrt {\left( { - 1 + 1} \right)} = 0$
So the value of the denominator is zero, when put the value of $x \to - 1$.
Here the both the numerators and the denominator are zero when $x \to - 1$, hence applying the L’Hospital’s rule to the limit \[\mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}{{\sqrt {\left( {x + 1} \right)} }}\] as shown :
\[ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \dfrac{{\dfrac{d}{{dx}}\left( {\sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} } \right)}}{{\dfrac{d}{{dx}}\left( {\sqrt {\left( {x + 1} \right)} } \right)}}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \dfrac{{\left( {0 - \dfrac{1}{{2\sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}\dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)} \right)}}{{\dfrac{1}{{2\sqrt {\left( {x + 1} \right)} }}\dfrac{d}{{dx}}\left( {x + 1} \right)}}\]
We know that the derivative of ${\cos ^{ - 1}}x = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}$, hence substituting it below:
\[ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \dfrac{{\left( {0 - \dfrac{1}{{2\sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}\left( {\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}} \right)} \right)}}{{\dfrac{1}{{2\sqrt {\left( {x + 1} \right)} }}\left( {1 + 0} \right)}}\]
$ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \dfrac{{\dfrac{1}{{2\sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)}}{{\dfrac{1}{{2\sqrt {\left( {x + 1} \right)} }}}}$
Now 2 gets cancelled in both the numerator and the denominator, and substituting the limit of $x \to - 1$:

$ \Rightarrow \mathop {\lim }\limits_{x \to - 1} \dfrac{{\dfrac{1}{{\sqrt {\left( {{{\cos }^{ - 1}}\left( { - 1} \right)} \right)} }}\left( {\dfrac{1}{{\sqrt {1 - {{\left( { - 1} \right)}^2}} }}} \right)}}{{\dfrac{1}{{\sqrt {\left( {\left( { - 1} \right) + 1} \right)} }}}}$
$ \Rightarrow \dfrac{{\dfrac{1}{{\sqrt \pi }}\left( {\dfrac{1}{{\sqrt {1 - 1} }}} \right)}}{{\dfrac{1}{{\sqrt {\left( {1 - 1} \right)} }}}}$
Here in the numerator and the denominator the expression $\sqrt {1 - 1} $ gets cancelled, as shown :
$ \Rightarrow \dfrac{1}{{\sqrt \pi }}$

$\therefore \mathop {\lim }\limits_{x \to - 1} \dfrac{{\sqrt \pi - \sqrt {\left( {{{\cos }^{ - 1}}x} \right)} }}{{\sqrt {\left( {x + 1} \right)} }} = \dfrac{1}{{\sqrt \pi }}$

Note:
Please note that in mathematics, more specifically in calculus, L’Hospital’s rule provides a technique to evaluate limits of indeterminate forms. Application of the rule often converts an indeterminate form to an expression that can be easily evaluated by substitution.